hdoj 5475 An easy problem 【线段树单点更新 + 区间乘积】
An easy problemTime Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 647 Accepted Submission(s): 380
Problem Description
One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a number. 2. divide X with a number which was multiplied before. After each operation, please output the number X modulo M.
Input
The first line is an integer T(
1≤T≤10 ), indicating the number of test cases.
For each test case, the first line are two integers Q and M. Q is the number of operations and M is described above. ( 1≤Q≤105,1≤M≤109 ) The next Q lines, each line starts with an integer x indicating the type of operation. if x is 1, an integer y is given, indicating the number to multiply. ( 0<y≤109 ) if x is 2, an integer n is given. The calculator will divide the number which is multiplied in the nth operation. (the nth operation must be a type 1 operation.) It's guaranteed that in type 2 operation, there won't be two same n.
Output
For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow, each line please output an answer showed by the calculator.
Sample Input
Sample Output
|
比赛时逆元卡死了，最后康总线段树KO了。
题意：有Q次操作，操作分两种。
1 y —— 表示当前结果乘以y。
2 n —— 表示当前结果除去第n步乘上的y。
不能直接逆元，中间会出问题。。。
思路：可以用线段树维护区间的乘积来表示每次操作的结果，初始化所有区间值为1。碰到操作1更新当前位置的值为y，碰到操作2，把第n个位置的值变为1。
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 100000+10
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define LL long long
using namespace std;
struct Tree
{
int l, r;
LL sum;
};
Tree tree[MAXN<<2];
int a[MAXN];
int Q, M;
void PushUp(int o){
tree[o].sum = tree[ll].sum % M * tree[rr].sum % M;
}
void build(int o, int l, int r)
{
tree[o].l = l;
tree[o].r = r;
tree[o].sum = 1;
if(l == r)
return ;
int mid = (l + r) >> 1;
build(lson);
build(rson);
}
void update(int o, int pos, int val)
{
if(tree[o].l == tree[o].r)
{
tree[o].sum = val%M;
return ;
}
int mid = (tree[o].l + tree[o].r) >> 1;
if(pos <= mid)
update(ll, pos, val);
else
update(rr, pos, val);
PushUp(o);
}
LL query(int o, int L, int R)
{
if(L <= tree[o].l && R >= tree[o].r)
return tree[o].sum % M;
int mid = (tree[o].l + tree[o].r) >> 1;
if(R <= mid)
return query(ll, L, R);
else if(L > mid)
return query(rr, L, R);
else
return query(ll, L, mid) * query(rr, mid+1, R) % M;
}
int main()
{
int t, k = 1;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &Q, &M);
build(1, 1, Q);
printf("Case #%d:\n", k++);
for(int i = 1; i <= Q; i++)
{
int op;
scanf("%d%d", &op, &a[i]);
if(op == 1)
update(1, i, a[i]);
else if(op == 2)
update(1, a[i], 1);
printf("%lld\n", query(1, 1, i) % M);
}
}
return 0;
}